Assuming the atmospheric pressure is 1 atm at sea level, determine the atmospheric pressure at Badwater (in Death Valley, California) where the elevation is 86.0 m below sea level.

Answers

Answer 1

Answer:

Atmospheric pressure at Badwater is 1.01022 atm

Explanation:

Data given:

1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa

Elevation (h) = 86m

gravity (g) = 9.8 m/s2

Density of air P = 1.225 kg/m3

Therefore pressure at bad water Pb = Pi + Pgh

Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)

Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa

hence:

Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm


Related Questions

what is Friction
short note on friction​

Answers

Answer:

Explanation:

Friction can be defined as a force that resists the relative motion of two objects when there surface comes in contact. Thus, it prevents two surface from easily sliding over or slipping across one another. Also, friction usually reduces the efficiency and mechanical advantage of machines but can be reduced through lubrication.

Generally, there are four (4) main types of friction and these includes;

I. Static friction.

II. Rolling friction.

III. Sliding friction.

IV. Fluid friction.

A 0.160kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.820m/s . It has a head-on collision with a 0.300kg glider that is moving to the left with a speed of 2.27m/s . Suppose the collision is elastic.
Part A
Find the magnitude of the final velocity of the 0.160kg glider. m/s
Part B
Find the direction of the final velocity of the 0.160kg glider.
i. to the right
ii. to the left
Part C
Find the magnitude of the final velocity of the 0.300kg glider. m/s
Part D
Find the direction of the final velocity of the 0.300kg glider.

Answers

Answer:

A) v_{f1} = -3.2 m / s,  B) LEFT , C) v_{f2} = -0.12 m / s,  D) LEFT

Explanation:

This is a collision exercise that can be solved using momentum conservation, for this we define a system formed by gliders, so that the forces during the collision are internal and the moment is conserved.

Let's use the subscript 1 for the lightest glider m1 = 0.160 kg and vo1 = 0.820 m / s

subscript 2 for the heaviest glider me² = 0.820 kg and vo2 = -2.27 m / s

Initial instant. Before the crash

          p₀ = m₁ v₀₁ + m₂ v₀₂

Final moment. After the crash

          p_f = m₁ v_{f1} + m₂ v_{f2}

          p₀ = p_f

          m₁ v₀₁ + m₂ v₀₂ = m₁ v_{f1} + m₂ v_{f2}

 as the shock is elastic, energy is conserved

         K₀ = K_f

         ½ m₁ v₀₁² + ½ m₂ v₀₂² = ½ m₁ [tex]v_{f1}^2[/tex] + ½ m₂ [tex]v_{f2}^2[/tex]

         m₁ (v₀₁² - v_{f1}²) = m₂ (v_{f2}² -v₀₂²)

let's make the relationship

         (a + b) (a-b) = a² -b²

         m₁ (v₀₁ + v_{f1}) (v₀₁-v+{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

let's write our two equations

         m₁ (v₀₁ -v_{f1}) = m₂ (v_(f2) - v₀₂)                                  (1)

         m₁ (v₀₁ + v_{f1}) (v₀₁-v_{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

we solve

         v₀₁ + v_{f2} = v_{f2} + v₀₂

we substitute in equation 1 and obtain

         M = m₁ + m₂

         [tex]v_{f1} = \frac{m_1-m_2}{M} v_o_1 + 2 \frac{m_2}{M} v_f_2[/tex]

         [tex]v_f_2 = \frac{2m_1}{M} v_o_1 + \frac{m_2-m_1}{M} v_o_2[/tex]vf2 = 2m1 / mm vo1 + m2-m1 / mm vo2

we calculate the values

         m₁ + m₂ = 0.160 +0.3000 = 0.46 kg

         v_{f1} = [tex]\frac{ 0.160 -0.300} {0.460} \ 0.820 + \frac{2 \ 0300}{0.460} \ (-2.27)[/tex]

         v_{f1} = -0,250 - 2,961

          v_{f1} = - 3,211 m / s

 

         v_{f2} = [tex]\frac{2 \ 0.160}{0.460} \ 0.820 + \frac{0.300 - 0.160}{0.460 } \ (-2.27)[/tex]

         v_{f2} = 0.570 - 0.6909

         v_{f2} = -0.12 m / s

now we can answer the different questions

A) v_{f1} = -3.2 m / s

B) the negative sign indicates that it moves to the left

C) v_{f2} = -0.12 m / s

D) the negative sign indicates that it moves to the LEFT

a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?​

Answers

Answer:

Explanation:

Speed= distance/time

Or time = distance/speed

According to your question

Speed=15m/s

and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s

D= 1.2km = 1.2×1000m =1200meter

Time = distance/ speed

1200/15 =80second

Or. 1min and 20 sec will be your answer.

A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the vehicles get stuck together. What is their velocity after the collision? A. 2.9 m/s east B. 1.6 m/s east m C. 2.6 m/s east D. 1.8 m/s east​

Answers

Answer:

Explanation:

This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is

[tex][m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a[/tex]  Filling in:

[tex][1200(4.5)+2100(0)]=[(1200+2100)v][/tex] which simplifies to

5400 + 0 = 3300v

so v = 1.6 m/s to the east, choice B

write the formulae of magnesium chloride and sodium sulfate ​

Answers

Answer:

Magnesium Chloride: MgCl2

Sodium Sulfate: Na2SO4

A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q

Answers

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N

Coulomb's law equation

F = Kq₁q₂ / r²

Where

F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart

Data obtained from the question Initial distance apart (r₁) =  rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =?

How to determine the final force

From Coulomb's law,

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Kq₁q₂ = constant

F₁r₁² = F₂r₂²

With the above formula, we can obtain the final force as follow:

F₁r₁² = F₂r₂²

1.2 × r² = F₂ × (2r)²

1.2r² = F₂ × 4r²

Divide both side by 4r²

F₂ = 1.2r² / 4r²

F₂ = 0.3 N

Learn more about Coulomb's law:

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You place an 8 kg ball on the top of your 2 cm^2 finger tip. Calculate the
PRESSURE. Show MATH, answer and unit.

Answers

Answer:

the pressure exerted by the object is 392,000 N/m²

Explanation:

Given;

mass of the object, m = 8 kg

area of your finger, A = 2 cm² = 2.0 x 10⁻⁴ m²

acceleration due to gravity, g = 9.8 m/s²

The pressure exerted by the object is calculated as;

[tex]Pressure = \frac{F}{A} = \frac{mg}{A} = \frac{8 \times 9.8}{2\times 10^{-4}} = 392,000 \ N/m^2[/tex]

Therefore, the pressure exerted by the object is 392,000 N/m²

chemical kinetics half lives

Answers

Answer and Explanation:

The half-life of a reaction is the time required for a reactant to reach one-half its initial concentration or pressure. For a first-order reaction, the half-life is independent of concentration and constant over time.

Two guitar strings, of equal length and linear density, are tuned such that the second harmonic of the first string has the same frequency as the third harmonic of the second string. The tension of the first string is 510 N. Calculate the tension of the second string.

Answers

Answer:

The tension in the second string is 226.7 N.

Explanation:

Length is L, mass per unit length = m

T = 510 N

Let the tension in the second string is T'.

second harmonic of the first string = third harmonic of the second string

[tex]2 f = 3 f'\\\\2\sqrt{\frac{T}{m}} = 3 \sqrt {\frac{T'}{m}}\\\\4 T = 9 T'\\\\4\times 510 = 9 T'\\\\T' = 226.7 N[/tex]

Explain whether the unit of work is a fundamental or derived unit

Answers

Answer:

answer here

Explanation:

the unit of work is fundamental unit because it doesn't depend on other units.

__________________

Thx

Answer:

The SI unit of work is joule (J)

Explanation:

Joule is a derived unit. ∴ unit of work is a derived unit

A ​12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall​ (above the fire​ truck) if the ladder makes an angle of with the horizontal

Answers

Complete Question

A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall​ (above the fire​ truck) if the ladder makes an angle of  

40° 16' with the horizontal.

Answer:

 [tex]d=8.01m[/tex]

Explanation:

From the question we are told that:

Length of ladder [tex]l=12.5m[/tex]

Angle [tex]\theta=40° 16'=20.26 \textdegree[/tex]

Generally the Trigonometric equation for distance d it goes up the wall is mathematically given by

 [tex]d=l sin \theta[/tex]

 [tex]d=12.5 sin 40.26[/tex]

 [tex]d=8.01m[/tex]

SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building

Answers

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

A 5 kg object is moving in one dimension along the x-axis with a speed of 2 m/s. An external impulse acts on the force causing the speed of the object to increase to 5 m/s. The impulse lasted for 3 s. What is the average net force (in N) exerted on the object

Answers

Answer:

The correct answer is "15 Kg.m/s".

Explanation:

Given values are:

Mass,

m = 5 Kg

Initial velocity,

u = 2 m/s

Final velocity,

v = 5 m/s

Now,

The magnitude of change in linear momentum will be:

= [tex]m\times (v - u)[/tex]

By substituting the values, we get

= [tex]5\times (5 - 2)[/tex]

= [tex]5\times 3[/tex]

= [tex]15 \ Kg.m/s[/tex]

An electric heater is madde of a wire of resistance 100π and connected to a 240v mains supply. Determine the power rating of the heater​

Answers

Answer:

Power = 576 Watts

Explanation:

The electrical power of an electric circuit can be defined as a measure of the rate at which energy is either produced or absorbed in the circuit.

Mathematically, electrical power is given by the formula;

[tex] Electrical \; power = current * voltage [/tex]

This ultimately implies that, the quantity (current times voltage ) is electrical power and it is measured (S.I units) in Watt (W).

Given the following data;

Resistance = 100 ohms

Voltage = 240 V

To find the power rating of the heater;

Power = V²/R

Where;

V is the voltage.

R is the resistance.

Substituting into the formula, we have;

Power = 240²/100

Power = 57600/100

Power = 576 Watts

) Efficiency of a lever is always less than hundred percent.​

Answers

Yes. Because it opposes the law of friction

I hope this helps.

Explanation:

Please mark me brainliest

A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 59.4 m/s and returns the shot with the ball traveling horizontally at 37.2 m/s in the opposite direction. (Take the direction of the ball's final velocity (toward the net) to be the +x-direction).
(a) What is the impulse delivered to the ball by the racket?
(b) What work does the racket do on the ball?

Answers

5 9 . 4

- 3 7 . 2

2 2 . 2

Explanation:

Use the algorithm method.

5 9 . 4

- 3 7 . 2

2 2 . 2

2 Therefore, 59.4-37.2=22.259.4−37.2=22.2.

22.2

22.2

A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by

Answers

The question is incomplete. The complete question is :

A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].

Determine the following:

(a) frequency of the motion

(b) period of the motion

(c) amplitude of the motion

(d) first time after t = 0  that the object reaches the position x = 2.6 cm

Solution :

Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].

Comparing it with the general equation of simple harmonic motion,

 x = A sin (ωt + Φ)

  A = 4.7 cm

  ω = 7.9 π

a). Therefore, frequency, [tex]$f=\frac{\omega}{2 \pi}$[/tex]

                                             [tex]$=\frac{7.9 \pi}{2 \pi}$[/tex]

                                             = 3.95 Hz

b). The period, [tex]$T=\frac{1}{f}$[/tex]

                        [tex]$T=\frac{1}{3.95}[/tex]

                            = 0.253 seconds

c). Amplitude is A = 4.7 cm

d). We have,

    x = A sin (ωt + Φ)

    [tex]$x_t=4.7 \sin (7.9 \pi t)$[/tex]

    [tex]$2.6 = 4.7 \sin (7.9 \pi t)$[/tex]

     [tex]$\sin (7.9 \pi t) = \frac{26}{47}$[/tex]

     [tex]$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$[/tex]

          Hence, t = 0.0236 seconds.

When an automobile moves with constant velocity the power developed is used to overcome the frictional forces exerted by the air and the road. If the power developed in an engine is 50.0 hp, what total frictional force acts on the car at 55 mph (24.6 m/s)

Answers

P = F v

where P is power, F is the magnitude of force, and v is speed. So

50.0 hp = 37,280 W = F (24.6 m/s)

==>   F = (37,280 W) / (24.6 m/s) ≈ 1520 N

Imagine you were given a converging lens and a meter stick and sent outside on a sunny day. In a few sentences, describe a method to measure, as accurately as possible, the focal length of the lens using only the lens, a meter stick, and your outside surroundings. Explain your reasoning

Answers

Answer:

the Sun is the object that is at a very great distance and the focus point of the sun's shape is the distance to the magnesia, this image is equal to the seal distance

Explanation:

The method for measuring the focal length of a lens is based on the use of the constructor's equation

        [tex]\frac{1}{f } = \frac{1}{p} + \frac{1}{q}[/tex]

where q and q are the distance to the object and the image respectively, f is the focal length.

If we place the object very far away (infinity) the equation remains

        [tex]\frac{1}{f} = \frac{1}{q}[/tex]

Therefore with this we can devise a means for measuring the Sun is the object that is at a very great distance and the focus point of the sun's shape is the distance to the magnesia, this image is equal to the seal distance

You are driving to the grocery store at 20 m/s. You are 150 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.

Required:
a. How far are you from the intersection when you begin to apply the brakes?
b. What acceleration will bring you to rest right at the intersection?
c. How long does it take you to stop?

Answers

Hi there!

a.

Use the formula d = st to solve:

d = 20 × 0.5 = 10m

150 - 10 = 140m away when brakes are applied

b.

Use the following kinematic equation to solve:

vf² = vi² + 2ad

Plug in known values:

0 = 20² + 2(150)(a)

Solve:

0 = 400 + 300a

-300a = 400

a = -4/3 (≈ -1.33) m/s² required

c.

Use the following kinematic equation to solve:

vf = vi + at

0 = 20 - 4/3t

Solve:

4/3t = 20

Multiply both sides by 3/4 for ease of solving:

t = 15 sec

A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is the spring constant of the bungee cord, assuming it has negligible mass compared to that of the jumper in N/m

Answers

Answer:

2.80N/m

Explanation:

Given data

mass m= 56kg

perios T= 11.2s

The expression for the period is given as

T=2π√m/k

Substitute

11.2= 2*3.142*√56/k

square both sides

11.2^2= 2*3.142*56/k

125.44= 351.904/k

k=351.904/125.44

k= 2.80N/m

Hence the spring constant is 2.80N/m

(c) The ball leaves the tennis player's racket at a speed of 50 m/s and travels a
distance of 20 m before bouncing.
(i) Calculate how long it takes the ball to travel this distance.
(1 mark)

Answers

Answer:

t=0.417s

Explanation:

After the ball hits the racket it is in freefall(assume air resistance as negligible)

so a=-g

use

x-x0=v0t+1/2at^2

Plug in givens

20=50t-4.9t^2

Solve quadratic equation using quadratic formula

t= 0.417 seconds, (the other answer is extraneous because it is too big because in 1 second, the ball travels 50 meters)

A typical incandescent light bulb consumes 75 W of power and has a mass of 20 g. You want to save electrical energy by dropping the bulb from a height great enough so that the kinetic energy of the bulb when it reaches the floor will be the same as the energy it took to keep the bulb on for 2.0 hours. From what height should you drop the bulb, assuming no air resistance and constant g?

Answers

Answer:

h = 2755102 m = 2755.102 km

Explanation:

According to the given condition:

Potential Energy = Energy Consumed by Bulb

[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]

where,

h = height = ?

P = Power of bulb = 75 W

t = time = (2 h)(3600 s/1 h) = 7200 s

m = mass of bulb = 20 g = 0.02 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]

h = 2755102 m = 2755.102 km

If a bale of hay behind the target exerts a constant friction force, how much farther will your arrow burry itself into the hay than the arrow from the younger shooter

Answers

Answer:

The arrow will bury itself farther by 3S₁

Explanation:

lets assume; the Arrow shot by me has a speed twice the speed of the arrow fired by the younger shooter

Given that ; acceleration is constant , Frictional force is constant

                    A₂ =   A₁

Vf²₂ - Vi²₂ / 2s₂  = Vf₁² - Vi₁² / 2s₁ ---- ( 1 )

final velocities = 0

Initial velocities : Vi₂ = 2(Vi₁ )

Back to equation 1

0 - (2Vi₁ )² / 2s₂ =  0 - Vi₁² / 2s₁

hence :

s₂ = 4s₁

hence the Arrow shot by me will burry itself farther by :

s₂ - s₁ = 3s₁

Note :  S1 = distance travelled by the arrow shot by the younger shooter

The force an ideal spring exerts on an object is given by , where measures the displacement of the object from its equilibrium position. If , how much work is done by this force as the object moves from to

Answers

Answer:

The correct answer is "1.2 J".

Explanation:

Seems that the given question is incomplete. Find the attachment of the complete query.

According to the question,

x₁ = -0.20 mx₂ = 0 mk = 60 N/m

Now,

⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]

⇒      [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]

⇒      [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]

By putting the values, we get

⇒      [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]

⇒      [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]

⇒      [tex]=1.2 \ J[/tex]

Match each term with the best description.

a. Tightly woven fabric used to smother and extinguish a fire.
b. Consists of absorbent material that can be ringed around a chemical spill until the spill can be neutralized.
c. Device used to control small fires in an emergency situation
d. Provides chemical. physical. Health, and safety information regarding chemical reagents and supplies

1. Spill containment kit
2. Safety Data sheet
3. Fume hood
4. Fire extinguisher
5. Fire blanket

Answers

Answer:

A - 5

B - 1

C - 4

D -2

Explanation:

I don't have one i just know...

The fire blanket is a tightly woven fabric. The spill containment kit consists of absorbent material. Fire extinguishers control small fires and the safety data sheet provides chemical, health, and safety information.

(a) The fire blanket is a blanket, which may be quickly thrown over a fire to snuff out the flames, and comprises fire-resistant materials.

Hence, option (a) matches with option (5)

(b) In order to contain a chemical spill, absorbent items like pads, socks, or booms are frequently included in spill containment kits.

Hence, option (b) correctly matches with option (1).

(c) A fire extinguisher is a tool used to put out small fires during emergencies.

Hence, option (c) correctly matches with option (4).

(d) A Safety Data Sheet (SDS) gives in-depth details regarding a specific chemical or chemical mixture. It provides information about the physical characteristics of the chemical, any potential risks, safe handling and storage practices, emergency response strategies, and more.

Hence, option (d) correctly matches option (2).

To learn more about Fire extinguishers, here:

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Cold air rises because it is denser than water, is this true?​

Answers

Answer:

true

Explanation:

im not sure please dont attack me

No,hot air rises cold air sinks

A study finds that the metabolic rate of mammals is proportional to m^3/4 , where m is the total body mass. By what factor does the metabolic rate of a 70.0-kg human exceed that of a 4.91-kg cat?

Answers

Answer:

The mass of human is 2898 times of the mass of cat.

Explanation:

A study finds that the metabolic rate of mammals is proportional to m^3/4 i.e.

[tex]M=\dfrac{km^3}{4}[/tex]

Where

k is constant

If m = 70 kg, the mass of human

[tex]M=\dfrac{70^3}{4}\\\\=85750[/tex]

If m = 4.91 kg, the mass of cat

[tex]M'=\dfrac{4.91^3}{4}\\\\=29.59[/tex]

So,

[tex]\dfrac{M}{M'}=\dfrac{85750}{29.59}\\\\=2897.93\approx 2898[/tex]

So, the mass of human is 2898 times of the mass of cat.

In a double-slit experiment, the slit spacing is 0.120 mm and the screen is 2.00 m from the slits. Find the wavelength (in nm) if the distance between the central bright region and the third bright fringe on a screen is 2.75 cm.

Answers

Answer:

[tex]\lambda=550nm[/tex]

Explanation:

From the question we are told that:

The slit spacing [tex]d_s=0.120mm=>0.120*10^{-3}[/tex]

Screen distance [tex]d_{sc}=2.0m[/tex]

Third Distance  [tex]X=2.75cm=>2.75*10^{-2}[/tex]

Generally the equation for Wavelength is mathematically given by

[tex]\lambda=\frac{Xd_s}{n*d_{sc}}[/tex]

Where

n=number of screens

[tex]n=3[/tex]

Therefore

[tex]\lambda=\frac{2.75*10^{-2}*0.120*10^{-3}}{3*2}[/tex]

[tex]\lambda=0.055*10^{-5}[/tex]

[tex]\lambda=550nm[/tex]

find out the odd one and give reason (length, volume, time, mass​

Answers

Answer:

Time

Explanation:

The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.

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